We can expand $\frac{1}{2} \ln(e^{2x} - 2x)$ into a Taylor series around $x = 0$. Let’s break this process into steps:

Step 1: Write the expression.

We are expanding: $f(x) = \frac{1}{2} \ln(e^{2x} - 2x)$

Step 2: Expand $e^{2x}$ as a Taylor series.

The Taylor series expansion of $e^{2x}$ around $x = 0$ is: $e^{2x} = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{4x^4}{3} + \dots$

Step 3: Substitute into $e^{2x} - 2x$.

Substitute the series into $e^{2x} - 2x$: $e^{2x} - 2x = \left(1 + 2x + 2x^2 + \frac{4x^3}{3} + \dots \right) - 2x = 1 + 2x^2 + \frac{4x^3}{3} + \dots$

Step 4: Take the natural logarithm.

To expand $\ln(1 + u)$, use the series: $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$ Here, $u = 2x^2 + \frac{4x^3}{3} + \dots$. Substituting:

= 2x^2 + \frac{4x^3}{3} - \frac{1}{2}(2x^2)^2 + \dots$$ ### Step 5: Simplify terms. Expand and collect terms up to desired order: $$\ln(e^{2x} - 2x) = 2x^2 + \frac{4x^3}{3} - 2x^4 + \dots$$ ### Step 6: Multiply by $$\frac{1}{2}$$. Finally, multiply by $$\frac{1}{2}$$: $$f(x) = \frac{1}{2} \ln(e^{2x} - 2x) = x^2 + \frac{2x^3}{3} - x^4 + \dots$$ ### Final Taylor Series: $$\frac{1}{2} \ln(e^{2x} - 2x) = x^2 + \frac{2x^3}{3} - x^4 + \dots$$ Let me know if you'd like more detailed steps or higher-order terms! --- ### Questions to explore: 1. How do we generalize the logarithmic expansion for more complex functions? 2. Why do Taylor series converge near $$x = 0$$ for analytic functions? 3. How does the $$e^{2x}$$ term dominate the expansion for large $$x$$? 4. What would the Taylor series of $$\ln(e^{2x} + 2x)$$ look like? 5. Can you visualize the approximation using a graph? ### Tip: Always check the radius of convergence for Taylor series to ensure the approximation is valid in your region of interest.